Write the rectangular equation (x-4)^2 + y^2=16 in polar form
x^2 - 8 x + 16 + y^2 = 16
or
x^2 - 8 x + y^2 = 0
let theta = T
x = r cos T
y= r sin T
r^2 cos^2 T - 8 r cos T + r^2 sin^2 T = 0
but cos^2+ sin^2 = 1
so
r^2 - 8 r cos T = 0
r = 8 cos theta
(x-4)^2 + y^2=16
x^2-8x+16+y^2=16
x^2-8x+y^2=0
since x^2+y^2=r^2,
r^2=8x
r^2=8rcos(theta)
r=8cos(theta)
I hope this helped
The above notes are true, but you know that
(x-4)^2 + y^2=16
is a circle of radius 4 with center at (4,0)
You also know that the circle r = 2a cosθ is a circle of radius a with center at (a,0) in x-y coordinates.
So, this one will be r = 8cosθ
To convert the given equation from rectangular form to polar form, we can use the following conversions:
x = r * cos(theta)
y = r * sin(theta)
We substitute these values into the given equation:
(x - 4)^2 + y^2 = 16
(r * cos(theta) - 4)^2 + (r * sin(theta))^2 = 16
Expanding the equation:
r^2 * cos^2(theta) - 8r * cos(theta) + 16 + r^2 * sin^2(theta) = 16
Since cos^2(theta) + sin^2(theta) = 1, we can simplify:
r^2 - 8r * cos(theta) + 16 = 16
Canceling out the common term on both sides:
r^2 - 8r * cos(theta) = 0
Now, we can factor out r to obtain:
r(r - 8 * cos(theta)) = 0
From this equation, we can conclude that either r = 0 or r - 8 * cos(theta) = 0.
If r = 0, it represents the origin in polar coordinates.
If r - 8 * cos(theta) = 0, we can solve for r to get:
r = 8 * cos(theta)
Therefore, the polar form of the given equation is:
r = 8 * cos(theta)